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A question

  1. Jul 11, 2005 #1
    are there two odd numbers [tex]x,y[/tex]
    that are the solutions of the equation [tex]15x^2+y^2=2^{2000}[/tex]
     
  2. jcsd
  3. Jul 11, 2005 #2
    For various reasons, the diophantine equation 15a + b = 2^2000 has the solutions

    a = 2^2000 + 2^2000 * t,
    b = -14*2^2000 - 15*2^2000 * t,

    where t is some integer. Now, if x^2 = a and y^2 = b (with x, y natural), then it's necessary (but obviously not sufficient) that both a and b be positive. This places some severe restrictions on t, in fact, if you try to solve the system

    a >= 0
    b >= 0

    you'll find that t = -1 is the only possibility, but then a = 0, which isn't odd.
     
  4. Jul 11, 2005 #3

    Hurkyl

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    I'd be curious to know those reasons... :confused:
     
  5. Jul 11, 2005 #4
    Damn it, it should be a = 2^2000 + t, b = -14*2^2000 - 15t... I messed up trying to distribute a multiplication over a parenthesis. This breaks the "solution". :(
     
  6. Jul 12, 2005 #5
    so i suppose the question remains open? :confused:
     
  7. Jul 12, 2005 #6

    Zurtex

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    Are you sure it has any solutions?

    I've messed about with this and found if there is a solution then y must be one of the 4 forms:

    [tex]y = 1 + 30c \quad \text{or} \quad y = 11 + 30c \quad \text{or} \quad y = 19 + 30c \quad \text{or} \quad y = 29 + 30c \quad c \in \mathbb{Z}[/tex]
     
    Last edited: Jul 12, 2005
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